Evaluate $\int\cos^3x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\cos x-\dfrac13\sin^3x+C$ (Choice B) B $\sin x-\dfrac13\sin^3x+C$ (Choice C) C $-\sin x+\dfrac13\sin^3x+C$ (Choice D) D $\sin x+\dfrac13\cos^3x+C$
Explanation: This has an odd power of $\cos x\,$. We will first factor out a $\cos x$ and use the identity $\sin^2x+\cos^2x=1\,$. $\begin{aligned} \int\cos^3x\, dx &= \int\cos x\cdot\cos^2x\,dx \\\\ &=\int\cos x(1-\sin^2x)\,dx \end{aligned}$ We can now use a $u$ -substitution where $u=\sin x$ and $du=\cos x\,dx\,$. $\int\cos x(1-\sin^2x)\,dx\Rightarrow\int(1-u^2)\,du$ The integration is now easy. $\int(1-u^2)\,du=u-\dfrac13u^3+C$ Now substitute back to get a result about the original variable $x$. $ u-\dfrac13u^3+C\Rightarrow\int\cos^3x\, dx =\sin x-\dfrac13\sin^3x+C$